Prove that the time of flight 'T' and horizontal range 'R' of a projectile are connected by the equation $g T^{2} = 2 R tan θ$

Open in App

Solution

As we know that time of flight $T = \frac{2 u S i n θ}{g}$ so $T^{2} = 4 u^{2} S i n^{2} θ / g^{2}$
And horizontal range is $R = 2 u^{2} S i n θ C o s θ / g$
Now on dividing $T^{2}$ by $R$ we get $T^{2} / R = 2 t a n θ / g$
or $g T^{2} = 2 R t a n θ$ ,proved.

Suggest Corrections

Similar questions

Q. A projectile is projected with some velocity at an angle

θ

from the horizontal so that its range is

R

and time of flight is

T

then

Q. A projectile has a time of flight

T

and range

R

. If the time of flight is double, what happens to the range?

If $T$ is the total time of flight, $h$ is the maximum height & $R$ is the range for horizontal motion, the $x$ and $y$ co-ordinates of projectile motion and time $t$ are related as:

Q. If

T

is the total time of flight,

h

is the maximum height and

R

is the range of horizontal motion, the

x

and

y

co-ordinates of projectile motion and time

t

α

given by

tan α

is equal to?

Join BYJU'S Learning Program

Prove that the time of flight 'T' and horizontal range 'R' of a projectile are connected by the equation gT2=2Rtanθ

As we know that time of flight T=2uSinθg so T2=4u2Sin2θ/g2And horizontal range is R=2u2SinθCosθ/gNow on dividing T2 by R we get T2/R=2tanθ/gor gT2=2Rtanθ,proved.

Prove that the time of flight 'T' and horizontal range 'R' of a projectile are connected by the equation $g T^{2} = 2 R tan θ$

As we know that time of flight $T = \frac{2 u S i n θ}{g}$ so $T^{2} = 4 u^{2} S i n^{2} θ / g^{2}$
And horizontal range is $R = 2 u^{2} S i n θ C o s θ / g$
Now on dividing $T^{2}$ by $R$ we get $T^{2} / R = 2 t a n θ / g$
or $g T^{2} = 2 R t a n θ$ ,proved.